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Chapter 1: Problem 8

In Problems 1-30, plot the graph of each equation. Begin by checking for symmetries and be sure to find all \(x\) - and \(y\)-intercepts. $$ y=3 x^{2}-2 x+2 $$

Short Answer

Expert verified
The vertex is \( (\frac{1}{3}, \frac{4}{3}) \), y-intercept is \( (0, 2) \), no real x-intercepts, symmetry line is \( x = \frac{1}{3} \).

Step by step solution

01

Identify the Type of Equation

The given equation \( y = 3x^2 - 2x + 2 \) is a quadratic equation. This means its graph will be a parabola.
02

Check for Symmetries

For quadratic equations in the form \( ax^2 + bx + c \), check for symmetry about a vertical line. The line of symmetry (axis of symmetry) can be calculated using the formula \( x = -\frac{b}{2a} \). Substituting \( a = 3 \) and \( b = -2 \), we get \( x = -\frac{-2}{2 \times 3} = \frac{1}{3} \). This is the axis of symmetry.
03

Find the Vertex

The vertex of a parabola in standard form \( y = ax^2 + bx + c \) can be found using \( x = -\frac{b}{2a} \). We already found that \( x = \frac{1}{3} \). Substitute \( x \) back into the equation to find \( y \): \( y = 3(\frac{1}{3})^2 - 2(\frac{1}{3}) + 2 = \frac{1}{3} - \frac{2}{3} + 2 = \frac{4}{3} \). Thus, the vertex is at \( \left( \frac{1}{3}, \frac{4}{3} \right) \).
04

Find the y-Intercept

The y-intercept is the point where the graph intersects the y-axis, which occurs when \( x = 0 \). Substitute \( x = 0 \) into the equation: \( y = 3(0)^2 - 2(0) + 2 = 2 \). Thus, the y-intercept is at \( (0, 2) \).
05

Find the x-Intercepts (Roots)

The x-intercepts are the points where \( y = 0 \). Set the equation \( 3x^2 - 2x + 2 = 0 \) and solve for \( x \). Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -2 \), and \( c = 2 \), calculate the discriminant: \( D = (-2)^2 - 4 \times 3 \times 2 = 4 - 24 = -20 \). Since the discriminant is negative, there are no real x-intercepts.
06

Sketch the Graph

Using the calculated vertex \( (\frac{1}{3}, \frac{4}{3}) \), y-intercept \( (0, 2) \), and the fact that there are no real x-intercepts, sketch the parabola. The parabola opens upwards (as the leading coefficient of \( x^2 \) is positive) and is symmetric about the line \( x = \frac{1}{3} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parabola Graphing
In mathematics, a quadratic equation like \( y = 3x^2 - 2x + 2 \) forms a curve called a parabola. Parabolas can open upwards or downwards. In our example, since the coefficient of \( x^2 \) (which is 3) is positive, the parabola opens upwards. This is crucial for determining the general shape during graphing.
To graph a parabola accurately, you need to consider several key features such as the vertex, axis of symmetry, and intercepts. These components help pinpoint the parabola's precise location and shape on the graph. Start by plotting the vertex and then sketching the curve's general direction based on whether it opens upwards or downwards.
Symmetry in Equations
Quadratic equations possess an interesting property: symmetry. This means if you fold the graph along a vertical line, the two halves would mirror each other perfectly. For our equation, the symmetry can be calculated using the formula \( x = -\frac{b}{2a} \).
Given the equation \( y = 3x^2 - 2x + 2 \), substituting \( a = 3 \) and \( b = -2 \), gives us \( x = \frac{1}{3} \). This line, \( x = \frac{1}{3} \), is known as the axis of symmetry. It's a crucial feature as it helps in graphing, allowing you to reflect points across this axis and ensuring the accuracy of your drawn parabola.
Vertex of a Parabola
The vertex is a point on the parabola that represents its highest or lowest point, depending on the parabola's orientation. For an upward-opening parabola like \( y = 3x^2 - 2x + 2 \), the vertex is the lowest point.
To find the vertex, use the axis of symmetry \( x = \frac{1}{3} \) and substitute it back into the equation to find \( y \). This gives: \( y = 3(\frac{1}{3})^2 - 2(\frac{1}{3}) + 2 = \frac{4}{3} \). Thus, the vertex is at \( (\frac{1}{3}, \frac{4}{3}) \). The vertex provides a starting point for sketching the parabola and is essential for understanding the direction and location of the curve.
X and Y Intercepts
Intercepts are points where the graph crosses the axes. The y-intercept is where the graph crosses the y-axis, which happens when \( x = 0 \). In the equation \( y = 3x^2 - 2x + 2 \), setting \( x = 0 \) gives \( y = 2 \), therefore the y-intercept is at \( (0, 2) \).
X-intercepts are where the graph crosses the x-axis, found by setting \( y = 0 \). However, for this quadratic, \( 3x^2 - 2x + 2 = 0 \), the discriminant \( -20 \) is negative, indicating no real x-intercepts exist.
These intercepts are vital as reference points when sketching the graph, providing insight into where the parabola interacts with the axes and highlighting further properties of the curve.

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